If first (r) objects are selected and then arranged in order then which formula for (^{n}P_r) is naturally obtained?
Correct answer: B
Each selected group can be arranged in (r!) ways. In exams treat permutation as selection followed by arrangement.
Mathematics
Derivations of formulas and their connections
सूत्रों की व्युत्पत्तियाँ और उनके पारस्परिक संबंध
In Class 11 Mathematics, this topic explains how the formulas in Permutations and Combinations are derived and how they are connected. Students learn the meaning of factorial notation, develop the formulas for nPr and nCr from counting principles, and understand why nPr = r! nCr and nCr = nC(n−r). The topic also shows when to use arrangements or selections, helping students apply these relationships logically instead of relying on memorized formulas.
TOPIC PRACTICE
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Hard · Level 1
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- (^{n}P_r=^{n}C_r+r!)
- (^{n}P_r=^{n}C_r\times r!)
- (^{n}P_r=\frac{^{n}C_r}{r!})
- (^{n}P_r=n!\times r!)
Hard · Level 1
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- Because arrangements of (r) objects are not counted
- Because (n-r) objects are selected
- Because all objects are identical
- Because repetition is allowed
Hard · Level 1
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- (^{n}P_r=\frac{n!}{r!})
- (^{n}P_r=\frac{n!}{(n-r)!})
- (^{n}P_r=\frac{(n-r)!}{n!})
- (^{n}P_r=\frac{r!}{n!})
Hard · Level 1
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- Arranging selected objects in order
- Choosing (r) and leaving (n-r) are equivalent decisions
- Selecting every object twice
- Counting all orders separately
Hard · Level 1
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- A fixed object is included or not included
- Arranging all objects first
- Multiplying only by (r!)
- Writing every selection in reverse
Hard · Level 1
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- (\frac{^{n}C_{r+1}}{^{n}C_r}=\frac{r+1}{n-r})
- (\frac{^{n}C_{r+1}}{^{n}C_r}=\frac{n-r}{r+1})
- (\frac{^{n}C_{r+1}}{^{n}C_r}=n-r)
- (\frac{^{n}C_{r+1}}{^{n}C_r}=r+1)
Hard · Level 1
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- (^{n}C_r=^{n}P_r\times r!)
- (^{n}C_r=\frac{^{n}P_r}{r!})
- (^{n}C_r=\frac{r!}{^{n}P_r})
- (^{n}C_r=^{n}P_r-r!)
Hard · Level 1
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- Because (L) appears twice
- Because (E) appears twice
- Because (V) appears twice
- Because all (5) letters are different
Hard · Level 1
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- (^{7}C_3)
- (^{7}P_3)
- (\frac{^{7}P_3}{7!})
- (^{3}C_7)
Hard · Level 1
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- Because the two posts are different
- Because the two posts are identical
- Because a player can be chosen twice
- Because order is ignored
Hard · Level 1
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- Because seating order is needed
- Because posts in the committee are different
- Because only a group is needed
- Because repetition is allowed
Hard · Level 1
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- Every position always has (n) choices
- Choices are (n), then (n-1), then decreasing
- There is only one choice
- Every object is repeated
Hard · Level 1
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- (^{n}C_r=^{n}P_r)
- (^{n}C_r=^{r}C_n)
- (^{n}C_r=^{n}C_{n-r})
- (^{n}C_r=r!)
Hard · Level 1
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- 2
- \(2!\)
- \(n\)
- \(\frac{1}{2!}\)
Hard · Level 1
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- There is one way to choose no object
- There is no way
- All objects must be chosen
- Order prevents (0!)
Hard · Level 1
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- There is exactly one way to choose all (n) objects
- Every object must be arranged differently
- No object is selected from (n) objects
- It is always equal to (n!)
Hard · Level 1
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- Selecting an unordered committee of r objects from n distinct objects
- Choosing r objects from n distinct objects and arranging them in r distinct positions
- Choosing r objects from n distinct objects when repetition is allowed
- Arranging all n distinct objects in a complete order
Hard · Level 1
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- \(4!\)
- \(\frac{12!}{8!}\)
- \(\frac{12!}{4!8!}\)
- \(\frac{1}{4!}\)
Hard · Level 1
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- (^{6}C_4)
- (^{6}P_4=^{6}C_4\cdot4!)
- (^{6}P_4=^{6}C_4+4!)
- (^{4}P_6)
Hard · Level 1
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- There are (n) independent choices for selecting one object
- Selecting one object gives (n!) arrangements
- Selecting one object gives ((n-1)!) choices
- There is no way to select one object
Hard · Level 1
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- Because the order of (1) object creates no different result
- Because (n=1) is necessary
- Because (r=0)
- Because repetition is allowed
Hard · Level 1
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- Because ordered pairs are divided by (2!) to become unordered pairs
- Because (n) is always even
- Because the two objects are identical
- Because (n-2) objects are selected
Hard · Level 1
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- Complement identity
- Pascal identity
- Product rule
- Factorial definition
Hard · Level 1
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- (4)
- (5)
- (6)
- (7)
Hard · Level 1
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- (3!)
- (12!)
- (15!)
- (\frac{1}{3!})
Why do we divide by (r!) while deriving (^{n}C_r=\frac{n!}{r!(n-r)!})?
Correct answer: A
In combinations order is not important so (r!) duplicate arrangements are removed. In exams divide when order is ignored.
If (^{n}P_r=n(n-1)\cdots(n-r+1)) then which factorial form is correct?
Correct answer: B
After (r) decreasing factors the remaining tail is ((n-r)!). In exams put the missing tail in the denominator.
The relation (^{n}C_r=^{n}C_{n-r}) is directly connected with which idea?
Correct answer: B
Choosing (r) objects is equivalent to not choosing (n-r) objects. In exams use complement selection to identify symmetry.
Pascal identity (^{n}C_r=^{n-1}C_r+^{n-1}C_{r-1}) comes from which counting split?
Correct answer: A
Make cases on one fixed object: include it or exclude it. In exams derive such identities by inclusion cases.
If (^{n}C_{r+1}=\frac{n-r}{r+1},^{n}C_r) then this relation is derived from which ratio?
Correct answer: B
Writing factorial forms and canceling common terms gives the ratio. In exams ratio method is fast for consecutive combinations.
From the derivation (^{n}P_r=^{n}C_r r!), what is the correct rearranged form of (^{n}C_r)?
Correct answer: B
A permutation includes (r!) orders for each combination. In exams divide ordered count by (r!) to get unordered count.
While finding distinct arrangements of the word (LEVEL), why does the formula (\frac{5!}{2!}) appear?
Correct answer: B
Interchanging identical (E)'s does not give a new arrangement. In exams divide by factorials of repeated identical items.
If (3) books out of (7) distinct books are to be placed on a shelf in order then which formula connection is correct?
Correct answer: B
Both selection and order are needed so (^{7}P_3=^{7}C_3\cdot3!). In exams use permutation when order matters.
Why is (^{8}P_2) used for choosing a captain and a vice-captain from (8) players?
Correct answer: A
Switching captain and vice-captain changes the outcome. In exams treat different roles as order important.
If a committee of (4) students is formed from (9) students then why is (^{9}C_4) sufficient?
Correct answer: C
Only membership is counted in a committee, not order. In exams use combination for a group without posts.
The formula (n!) for arranging (n) distinct objects comes from which product principle idea?
Correct answer: B
After each chosen object available choices decrease. In exams write decreasing products for distinct arrangements.
If (^{10}C_3=^{10}C_7), this is an example of which formula connection?
Correct answer: C
This is the symmetry of choosing (r) and leaving (n-r). In exams use this relation when opposite indices sum to (n).
If \(^{n}P_2=^{n}C_2\cdot k\), what is the value of (k)?
Correct answer: B
The relation between permutations and combinations is \(^{n}P_r=^{n}C_r\cdot r!\). Here \(r=2\), so \(^{n}P_2=^{n}C_2\cdot 2!\), giving \(k=2!\). Note that \(2\) is numerically equal to \(2!\), so listing both as separate options would create two correct answers; always check equivalent expressions in MCQs. Exam tip: order matters in permutations, so each combination is multiplied by \(r!\).
The derivation of (^{n}C_0=1) is connected with which basic counting idea?
Correct answer: A
The empty selection is also a valid selection. In exams remember both (0!) and empty choice as (1).
Which is the correct reasoning for (^{n}C_n=1)?
Correct answer: A
When all objects are chosen the selection is fixed. In exams quickly identify extreme cases (^{n}C_0) and (^{n}C_n).
Which of the following situations is represented by the expression \(\binom{n}{r}r!\)?
Correct answer: B
First choose r objects in \(\binom{n}{r}\) ways, then arrange them in \(r!\) orders. Thus it represents selection followed by ordering; a committee has no order. Exam tip: multiply by \(r!\) whenever positions or order matter.
If \(^{12}P_4=^{12}C_4\cdot x\), what is the correct value of (x)?
Correct answer: A
The relation is \(^{n}P_r=^{n}C_r\cdot r!\), because the \(r\) selected objects can be arranged in \(r!\) orders. Here \(r=4\), so \(x=4!\). \(\frac{12!}{8!}\) is \(^{12}P_4\) itself, while \(\frac{12!}{4!8!}\) is \(^{12}C_4\) itself. Exam tip: in the relation between permutations and combinations, the multiplying factor is always \(r!\).
(4)-letter codes are formed from (6) distinct letters without repetition. Which formula connection is correct?
Correct answer: B
Changing order changes the code. In exams usually use permutation relation for codes or passwords.
Which statement correctly explains the derivation of (^{n}C_1=n)?
Correct answer: A
In single selection order does not arise and choices are (n). In exams both (^{n}C_1) and (^{n}P_1) equal (n).
Why are both (^{n}P_1=n) and (^{n}C_1=n) equal?
Correct answer: A
One object can be arranged in (1!) way. In exams permutation and combination give the same value for (r=1).
If (^{n}C_2=\frac{n(n-1)}{2}), where does the denominator (2) come from?
Correct answer: A
(AB) and (BA) are the same pair so repetition is removed. In exams (\frac{n(n-1)}{2}) is often useful for pairs.
Which identity can be written as (^{n}C_r+^{n}C_{r-1}=^{n+1}C_r)?
Correct answer: B
This is a shifted form of Pascal identity. In exams add adjacent lower-row combinations to form the upper-row combination.
If (^{14}C_r=^{14}C_{r+4}), what is the value of (r)?
Correct answer: B
The direct answer is option B: r = 5. The key idea is the symmetry property of combinations: for a fixed n, \\(^{n}C_k=^{n}C_{n-k}\\). Thus, two combination values with the same upper number are equal when their lower numbers are complementary, so r+(r+4)=14. Now solve step by step: r+r+4=14; 2r+4=14; 2r=10; r=5. Therefore, \\(^{14}C_5=^{14}C_9\\), and r+4=9, confirming the result. Option A, r=4, would give lower indices 4 and 8; these are not complementary because 4+8=12, not 14, so the values are not equal. Option B, r=5, gives 5 and 9, whose sum is 14, so it is correct. Option C, r=6, gives 6 and 10; their sum is 16, so it does not satisfy the symmetry condition. Option D, r=7, gives 7 and 11; their sum is 18, so it is also wrong. Remember: when \\(^{n}C_a=^{n}C_b\\) in this standard situation, use a+b=n, then solve carefully.
When connecting (^{15}P_3) with (^{15}C_3), which factor is used?
Correct answer: A
First choose (3) objects and arrange them in (3!) ways. In exams remember (^{n}P_3=^{n}C_3\cdot3!).
QUIZ COMPLETE